Question

A body travels 200 cm in the 1st 2 seconds & 220 cm in the next 5 seconds . Calculate the velocity of the body at the end of the 7th second .

Answer 1

hiii mate here is the answer

Acceleration of body is

a = 2S / t²

= (2 × 200 cm) / (2 s)²

= 100 cm/s²

Use equation of motion

v² - u² = 2aS

v = √(u² + 2aS)

= √[(0 m/s)² + (2 × 100 cm/s² × (200 cm + 220 cm))]

= 289.83 cm/s

≈ 3 m/s

∴ Velocity of body at the end of 7th second is approximately 3 m/s

hope it may help you

Answer 2

let initial velocity = u & accleration is a then

distance travelled in 4 sec = S6

using , s=ut + at2/2

  S6 = 4u + a36/2

     =  6u+18a

distance covered in 2 sec = S2 = 2u + 4a/2

      =2u+2a = 200   (given)

   u+a = 100 .........1

now distance covered in 2 to 6 sec is given 220cm

    S6-200 = 220

   6u + 18a = 420       

    u+3a = 70.................2

from 1& 2

a = -15 & u=115

velocity after 7 sec = V7

       V7 = u + at

           =115 - 15*7=10cm/sec

:)